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Potential energy difference as a gate to perpetual energy transfers

  1. Jul 16, 2003 #1

    Take a look to this:


    Basically, it's a method to trigger an enormous energy transfer, using a little energy input. A difference on potential energy between two points allows opposite forces (making a virtual zero) to create a real transference. Changing the little potential levels, these transferences happens once and again, converting potential energy of an elastic mechanism into potential energy of water weight, then into kinetic energy of water transfer, then into potential energy of the same water weight than before, and finally into potential energy into another elastic mechanism. This transfer is modulated by the little potential difference created on water levels without changing the water weight.

    Tell me what do you think.
  2. jcsd
  3. Jul 16, 2003 #2


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    Cala, Cala, Cala....

    I have but one thing to say... Build it.

    You have totally neglected any system losses. So if you have a complete frictionless system this may well oscillate for a long period of time. Of course as soon as you extact any work from the system it will grind to a halt.

    Redo your models throwing in some realistic friction forces and see what happens.
  4. Jul 16, 2003 #3
    Agreed. Might work in "ideal" circumstances, but more than likely not reality.

    How are those tubes going to expand and contract .02m?

    What about energy lose to the system?

    Many things that would not make this work.
  5. Jul 16, 2003 #4
    In all that time posting on Physicsforums, i never heard about any system allowed to give more output than input, not even under ideal conditions.

    Ideal conditions allow the systems to be as much as input=output...

    So here we have the first system with overunity allowed under ideal conditions!. (...i'm being sarcastic here )

    Well, that is true to closed systems, but if you consider this is an open system, then it can have more output than input not only on ideal conditions.

    If under ideal conditions you think the system can work, what rule of the universe makes you say that all the losses not taked into account will make the system eficiency go down from 257% to be under 100%?

    Maybe it could go from 257% to 145%, or to 178%...
  6. Jul 16, 2003 #5
    Second Law of Thermodynamics
  7. Jul 16, 2003 #6
    HA; HA; Not quite, cala. I challenge you to think again about the real definition of gravitational potential energy so you can see how this author dupped you into believing something erroneous about potential energy of 'water levels'. It's not just friction losses, etc. as everyone seems to be guessing.

    Its simply a matter of the correct definition of gravitational potential energy.

    Well, you say, he's using P.E.= mgh, right? Of course, but he is using 'h' in the formula to be the 'upper level of the water'. NOT!

    'h' in the formula is suppose to be the height of the CENTER OF MASS of the water! Not once did he calculate the center of mass of the water in the container. He merely calculates the difference in potential energy of the two containers using the upper 'level of the water'.
    How foolish' . Actually, pretty smart deception, seeing everybody went along with it .

    In order to calculate potential energy difference correctly, you would need to figure the center of mass of the water in EACH container for EACH of the 8 steps.
    Doing it thus will give quite a different story.

    In fact, if you're quick enough you can simply eyeball the center of mass of each container in the graphic as it is moving and you'll be able to see a huge change in the center of mass during each cycle. Each drop in center of mass must be replenished by some greater motive force EACH cycle to push the center of mass back up against gravity, (and it can't just be the weight of the water on the other side of the spring!)

    Last edited: Jul 16, 2003
  8. Jul 17, 2003 #7
    CREATOR, you missed the thing...

    I'm not using gravitational definition. There is something called HIDRAULIC POTENTIAL:

    Imagine a colum of water that begins at z height, and is w long.
    The hidraulic potential is defined as energy per unit mass: g*h=g(z+w)

    So, my calculus of the potential energy of 20 Kg of water at a height difference of 0.17 than the previous situation is quite right.

    Here is better explained:
    http://web.usal.es/~javisan/hidro/temas/T090.pdf [Broken]
    (sorry, in spanish)

    Also, once the transference is triggered, the proccess maintain itself, until a new equilibrium point happens. But we don't let the system rise this equilibrium point, changing the little potential difference of the water levels on the tanks to trigger the whole system tendency, causing the system to try to rise the other side equilibrium point.

    The energy input concerns only to the water tanks, but the effects of this little change of water levels when you connect the tanks make the potential energy stored on the elastic mechanism to begin to transfer. The energy input doesn't know nothing about any other energy stored on elastic mechanisms. The energy input is only related to tanks, but once we connect them, the elastic mechanism reacts to the energy input that caused the water level difference, trying to rise a new equilibrium that it will never rise.

    The energy input is just to "wake up" the "sleeping dragon". (What?? ...hehe)

    I will try to explain myself better, if you don't understand quite yet.
    Last edited by a moderator: May 1, 2017
  9. Jul 17, 2003 #8
    I say it too. There's a 1000000$ award on the net for any device that can give more energy than you put in. I'm sure a simple search would give you the site... :wink:
  10. Jul 17, 2003 #9
    Two things:

    So you are using a a potential energy to creat and produce this electrical energy. So the question is: Where does the turbine (generator) gain its energy from? The flow fo the water. Since the water is flowing, you now have turned potential energy into kenetic energy. This kenetic energy hits the turbine and loses some of it's energy to the turbine to create motion of the turbine. Thus when you get to the other side, you're systems energy has lost some of it because it has down the line become electrical energy.

    Have you looked at entropy of this system? I can hear Gibbs rolling in his grave right now. :-) j/k

  11. Jul 17, 2003 #10
    I'll try to explain how it works, and how it gets more energy than the needed:

    Imagine one of the tanks is full, and the other near empty, but both are at the same water level.
    In this situation, the elastic mechanism of the full tank is compressed and in opposition to the huge weight of the tank, and the elastic mechanism of the empty tank is expanded, and opposing to that little weight.

    When the water level is the same, the system is on static equilibrium.
    Now, we close the interconnection of the tanks, and change the water level of each tank. The static equilibrium is maintained, because the weight of the water on each tank doesn't change.

    We change the water levels by increasing or reducing the radius of the tanks. The energy needed doesn't interfere, doesn't oppose and has nothing to do with the energy stored on the elastic mechanism (remember that the weight doesn't change).

    So the input is a little energy, that changes the water level. Nothing more is afected.

    Now, we open the tanks interconnection. Now, the water level on each tank is different, so this causes a water transfer from the high level to the lower one. But this transfer not only reduce the water level, it also affects the weight of the tanks, so now, this variation on weight affects the elastic mechanisms stored energy. The input only affected the water level, not the weight, but the water transfer affects the level and the weight of the water, so the energy stored on the elastic mechanisms is transferred also. Now, the tanks and the elastics want to rise a new situation static equilibrium, but the energy output is not only by the water transfer, we have to add the potential energy transfer induced into the elastic mechanisms.

    The input is only changing water level, and doesn't affect the elastics, but the change made and the water transfer affects the weight, and then the elastics. The energy of elastics is liberated at this moment, and is not present when we have to apply the energy input.
  12. Jul 17, 2003 #11


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    You can't, because it won't.
    That is gravitational potential energy.

    All that system of yours does is transfer the same energy back and forth between the two tanks and the elastic band. There is no way for it to generate more energy. It is a closed system.
    Here is the extra energy you are not accounting for. Changing the radius of the tanks requries (or gives back) ENERGY because you are changing the gravitational potential energy of the water in the tank. You MUST account for this energy in your calculations.

    Again, you have a system that if you lubricate it well will oscillate for a little while before grinding to a halt. It does NOT generate energy.

    If you don't believe us, build it (or become a mechanical engineering major).
  13. Jul 17, 2003 #12
    russ, as you said: "all that the system does is tranfer the same energy back and forth between the tanks and the elastics"

    That's it. The energy stored on one elastic is passed to the tank, and it transfers it to the other side. Later, we need to apply less energy to reverse the proccess to be to the other side. Is the same potential energy all time, but there is a real moment of transfer of this energy from one elastic to the other through the water mechanism. potential energy of water is more on one side than on the other in all the transfer proccess, The elastics potential energy is the "source" that brings energy to the water transference to maintain the water potential difference, so the potential energy of the elastics can pass completely from one side to the other during the transfer. Then the real thing is that there is a potential difference, and the energy difference is maintained, although there is a real transference of water, so there is a "latent" energy that takes place on the transfer, and is stored on the other side. When the transfer finish, all the energy passed to this side, then we have to change the potential difference on water to re-use the energy supply stored on the elastic to the other side.
  14. Jul 17, 2003 #13


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    No. That is a direct violation of the first law of thermodynamics.
  15. Jul 17, 2003 #14

    I didn't miss a thing...YOU, cala, obviously missed something though; its called CORRECT physics.(see below).

    That exactly why you're having all the problems. If you want people to believe you, you need to use standard physics definitions.

    OK, if you want to use that; its merely g x height difference.

    Wrong again. The only thing you calculated here is the potential energy difference of the small amount of water which is above or below the upper level water line (0.17 height difference)! What about the potential energy of THE REST OF THE WATER!? DUH!

    YOU HAVE TOTALLY IGNORED THE LARGE MASS OF WATER BELOW THAT LINE whose center of mass is continually changing. Can't you see that? You may WANT to ignore the center of mass of that large amount of water, BUT standard physics says you CAN'T ignore it since it IS this center of mass change that BY DEFINITION determines how the potential energy changes.

    Again, as I stated at first, (and as Russ tried also to point out), you cannot ignore the center of mass change of the water (BELOW the level line) since that IS WHAT DETERMINES the potential energy transfer.
    Simply trying to deal with ONLY the difference AT THE TOP LEVEL IS NOT THE WHOLE STORY! You need to learn physics.....you must account for ALL the potential energy of ALL the water, NOT just the upper level change. If you are NOT WILLING to do that then it is quite obvious where the real problem lies, that is, in YOUR UNDERSTANDING of physics.

    I suggest YOU wake up, cala, [zz)]before trying to invoke some imaginary dragon.

    Last edited: Jul 17, 2003
  16. Jul 17, 2003 #15
    Imagine JUST the two water tanks, and forget the elastics.
    Imagine the two tanks are at the same level.
    Now, you reduce radius of one tank from 0.22 to 0.2 to lift 20Kg of water 0.17m.

    Now, you let the water pass from one tank to the other, moving a turbine in the proccess, until the water levels become equal, and the proccess stop. The total transfer of water is about 10Kg starting at 0.17m level difference.
    In this case, the energy extracted of the water flow is aprox the same you needed to lift the water (PE to KE conversion)

    Now, with the elastics, the energy needed to lift the same 20Kg of water 0.17m is the same as before, but you get much more a big transference of water, at aprox the same water levels.
  17. Jul 17, 2003 #16
    This is exactly the point Cala. Let me try explaining it another way myself:

    Take your two tanks like you have said and then compress the one. You've now used your energy to create a potential. So the Compression energy is now equal to the energy of the new water height.

    Now when the water flows out you will then spin a turbine and create electrical energy. (of course this would be at some lose that is not calculated in the theoretical data.) Then the water fills the other tank and then needs to compress itself.

    This is where the data has a problem.

    This is where you have neglected to find energy that is used. It does take energy to expand that tank. The energy required to expand (and compress the other) would be equal to the origonal potential created.(assuming complete transfer of electrical energy to the expansion/compression). You cannot ignore this, and this is what makes the data look much better.

    Imagine two ballons filled with water connected by a straw. If I compress one side of this system, then the other side will fill with a little more water. The energy put into your system is the potential. Because when i let go, the water will come right back in. Thus only giving me the amount of energy back out of the system. With a theorectical "turbine" in between to drive a electric generator, then you have to increase the amount of energy you push into this system (because of friction of the turbine). Thus it produces a little amount of energy because most of it goes into the other ballon. So the other ballon releases its potential and spin the turbine the opposite direction (i'm not sure if this is going to create the same power in the same with in both directions, you'll need to ask a person more familiar with AC. Though to put the ballon back into the state to create a potential again will take the same amount of energy which would require all of th energy comming from the electric generation.

    Now in the real world. you might get a system like this to run for a little while, but it will stop, due to friction, electric loses, and other entropy type things.

  18. Jul 18, 2003 #17
    The energy input takes place on a static situation, where the energies of the elastics are fully balanced with the weight of the tanks, making a virtual zero. So the energy input is the same as if the elastics were not there.

    But once the water transfer begins, the equilibrium of water weight and elastic tensions brakes, and then, the energy of the elastics is liberated, making work as the water flow. The energy output takes into account the effects of the variation of the elastics huge stored energy.

    We are using the difference of potential energy of water as energy output, but the transfer of water doesn't reduce this potential energy difference due to the effects of equilibration of weight and elastic tensions stored before, so it's as if we had a big tank that doesn't reduce it's level once created the potential difference with little energy

    If no elastics were there, the transfer of water will happen until the water levels being equal. The energy extracted will be the same needed to create the potential difference on the static situation. But there are elastic potentials in equilibrium with weight of water, that de-equilibrates with water flowing, and thus this inner elastic energy not present on the input stage helps the water to maintain the levels, and then allowing the transference of water to maintain, extracting more energy from the dinamyc energy output situation (where the energy of the elastics counts) than the needed on the static energy input situation (were the elastics doesn't count).
  19. Jul 18, 2003 #18


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    Sorry, cala. All energy inputs "count."

    I'm reminded of the (true) story about a guy who claimed to have invented a flywheel that gives back more energy than it took to run. He had a guy make the flywheel spin with a handcrank, then turned on a tiny electric motor, which kept the flywheel going. Then he disconnected the motor and connected a generator and got the flywheel to produce power until it stopped. The power he got out was greater than the power the motor was putting in to keep it running.

    See any problems with that?

    1. He ignored the input energy from the big burly man with the hand crank (which is ALL of the energy he got out of the system, minus losses).
    2. He didn't understand the difference between power and energy.

    This is also the same mistake the cold fusion guys made. Being chemists, not physicists (that shouldn't be an excuse, but I've heard it used), they didn't fully understand the first law of thermodynamics and the relationship between power and energy. They input power into their experiment for DAYS and then got a short, violent output power. Eureka - high output power, low input power. But oops - multiply by time and you find the ENERGY input was higher than the energy output.

    Cala, you are ignoring the input energy of getting the system started. And you are actually adding that energy back into the system every time the bands stretch, ignoring the fact that that energy is gone - it went out the wire from the generator. You actually have to manually stretch the rubber band each time. By doing that, you end up with a positive energy output from each oscillation, equal to the amount of energy (minus losses) that it took for you to manually stretch the rubber band each time.

    This is in addition to the fact as pointed out before, that you are ignoring the change in energy of the tank as its geometry changes.

    Cala, as someone else said before: build it. Hopefully once you see it doesn't work, you will indeavor to learn WHY. Though admittedly, people driven enough to acutally build a perpetual motion machine are also often deluded enough to believe the concept is still sound even after they see their device fail. Just have to work out the "kinks." As if the first law of thermodynamics were a "kink" to be worked out.
    Last edited: Jul 18, 2003
  20. Jul 18, 2003 #19

    What do you mean with "manually stretch the rubber bands"???

    The elastic or hidraulic mechanism respond to the weight of the water. To stretch them, you only have to put the water into the tank. The existence of the water is what compress the elastics. I have to do no work to stretch them, the water weight works for me.
  21. Jul 18, 2003 #20
    Also, when i say "the elastics energy doesn't count into static situation" (when you introduce the energy to change the water levels)i want to say the elastics stored energies is on opposition to the weight of the water. reduce the radius doesn't interfere the weight of the water, so the elastic doesn't care, but the water level is increased, and if you let the transfer of water happen, the elastic will react, and will contribute to the transference to keep going.
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